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This question already has an answer here:

In the book which I'm reading now,

the author says, in an arbitrary Unique Factorization Domain $R$,

$(a_0, a_1, ..., a_d) = (1)$ and $gcd(a_0, a_1, ..., a_d) = 1$, where each $a_i \in R $, is not necessarily equivalent.

I know the first one implies the seconed one, but in my native thought, the second one is seemed to imply the first one.

To disprove this, I'd like to see some example that in some UFD $R$, $(a_0, a_1, ..., a_d) \neq (1)$ but $gcd(a_0, a_1, ..., a_d) = 1$. But I can't come up with such domain $ R$.

Could you give me such example?

Thank you for kind answer in advance.

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marked as duplicate by Bill Dubuque abstract-algebra Aug 6 at 13:10

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After typing up this question, I came up with some example.

In $\mathbb Z[x]$, it's a well-known fact that $(2,x)$ is not a principal ideal but maximal ideal, since $\mathbb Z[x] /(2,x) \cong \mathbb Z/2Z$, where $\mathbb Z/2Z$ is a field.

So $(2,x) \neq 1$ in $\mathbb Z[x]$. But there is no element dividing $2$ and $x$ in $\mathbb Z[x]$ other than $1$.

This means that $(2,x) \neq 1$ but $gcd(2,x) = 1$, which is what I'd like to show.

Is this right? Any comment will be appreciated.

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    $\begingroup$ Yes, that works (+1). $\endgroup$ – quasi Aug 6 at 4:42
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Let $R=\mathbb{Q}[x,y]$.

Then $R$ is a UFD and $\gcd(x,y)=1$, but $(x,y)\ne (1)$.

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